---
title: "Check y(t) = x(t-2) + x(2-t)"
subject: "Data Communication"
chapter: 2
key: "semester_5/data_communication:check-xt-2-plus-x2-t"
source: https://www.bctengineer.com/q/semester_5/data_communication/check-xt-2-plus-x2-t
site: BCT Engineering
---

# Check y(t) = x(t-2) + x(2-t)

Data Communication · Chapter 2: Signals and Systems

- Asked in: 2074 Asoj, 3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/check-xt-2-plus-x2-t
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-2

## Question

Check whether the following system is linear, time invariant and causal or not: y(t) = x(t-2) + x(2-t)

## Short answer

The system y(t)=x(t-2)+x(2-t) is linear but time-variant due to the time reversal in x(2-t). It is non-causal because x(2-t) depends on future inputs (e.g., at t=0 it uses x(2)).

## Answer

y(t) = x(t - 2) + x(2 - t)

#### Linearity
For x = a·x₁ + b·x₂, y = a[x₁(t-2) + x₁(2-t)] + b[x₂(t-2) + x₂(2-t)] = a·y₁ + b·y₂. The system is **linear**.

#### Time invariance
Delay the input by t₀: x₁(t) = x(t - t₀). Then

y₁(t) = x₁(t - 2) + x₁(2 - t) = x(t - t₀ - 2) + x(2 - t - t₀)

Delay the output by t₀: y(t - t₀) = x(t - t₀ - 2) + x(2 - (t - t₀)) = x(t - t₀ - 2) + x(2 - t + t₀)

The second terms differ (x(2 - t - t₀) against x(2 - t + t₀)), so the system is **time-variant**. The time reversal in x(2 - t) causes this.

#### Causality
x(t - 2) is a past value (fine), but x(2 - t) at t = 0 is x(2), a future value; for any t < 1 it uses a future input. So the system is **non-causal**.

**Answer: linear, time-variant, non-causal.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Is y(t) = x(t-2) + x(2-t) linear?
   A. Yes, it is a weighted sum of shifted/reversed inputs
   B. No, because of the time reversal
   C. No, because of the shift by 2
   D. Only if x(t) is even
   Answer: A. Scaling and adding inputs gives a*y1 + b*y2, so superposition holds. (Key point: Linearity)

2. Why is the system time-variant?
   A. The shift x(t-2) changes the amplitude
   B. Delaying the input gives x(2-t-t0) but delaying the output gives x(2-t+t0)
   C. Because the system has two terms
   D. The output is unbounded
   Answer: B. The time-reversed term moves the opposite way when the input is delayed, so the two results differ. (Key point: Time variance)

3. Why is the system non-causal?
   A. x(t-2) uses a future value
   B. It is not linear
   C. x(2-t) at t = 0 is x(2), a future input value
   D. It is time-variant
   Answer: C. For t < 1, 2-t > t, so the output uses future input. (Key point: Causality)

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